PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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Frobenius groups and fixed point free actions

Statement

Let $N$ and $H$ be subgroups of a finite group $G$ with $G = N ⋊ H$ , $1 < N$ and $1 < H$ , and let $H$ act on $N$ by conjugation, $h \cdot n : = h n h^{- 1}$ .

If $G$ is a Frobenius group with complement $H$ (so that $N$ is its Frobenius kernel), then every $1 \neq h \in H$ fixes only the identity of $N$ : the conjugation action of $H$ on $N ∖ {1}$ is free.
Conversely, if the conjugation action of $H$ on $N ∖ {1}$ is free, then $H$ is a Frobenius complement of $G$ .

Facts & Assumptions

Given: A finite group $G$ with subgroups $N, H$ such that $N ⊴ G$ , $G = N H$ and $N \cap H = {1}$ , with $1 < N$ and $1 < H$ , and the conjugation action $h \cdot n = h n h^{- 1}$ of $H$ on $N$ .

[F1]

$N ⊴ G$ , $G = N H$ , $N \cap H = {1}$ , and $H$ is called a complement to $N$ ; these are exactly the internal-semidirect-product conditions (An internal semidirect product and a complement to a normal subgroup).

[F2]

$N ⊴ G$ means $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

[F3]

A subgroup ${1} < H < G$ is a Frobenius complement exactly when $H \cap g H g^{- 1} = {1}$ for every $g \notin H$ (Frobenius complement and frobenius group).

[F4]

For a finite Frobenius group with complement $H$ the kernel set $N$ is normal and $G = N ⋊ H$ with $N \cap H = {1}$ and $∣ N ∣ = [G : H]$ (Frobenius semidirect product decomposition, Frobenius kernel theorem).

[F5]

$H$ and $N$ are subgroups: each contains the identity, is closed under products, and is closed under inverses (Subgroup).

Proof

technique · direct

1.1

Suppose first that $G$ is a Frobenius group with complement $H$ , so that $N$ is its Frobenius kernel, $N ⊴ G$ and $N \cap H = {1}$ by [F4]. Let $1 \neq h \in H$ and $1 \neq n \in N$ satisfy $h \cdot n = n$ , that is $h n h^{- 1} = n$ . Then $h n = n h$ and therefore $h = n^{- 1} h n \in H \cap n^{- 1} H n$ .

F2F4givenalgebra

1.2

Suppose conversely that the conjugation action of $H$ on $N ∖ {1}$ is free. Let $g \in G ∖ H$ . By [F1] write $g = n h$ with $n \in N$ , $h \in H$ ; if $n = 1$ then $g = h \in H$ , so $n \neq 1$ . Conjugating, $g H g^{- 1} = n h H h^{- 1} n^{- 1} = n H n^{- 1}$ , because $h H h^{- 1} = H$ : thus $H \cap g H g^{- 1} = H \cap n H n^{- 1}$ .

F1F5assume-hypalgebra

1.3

Let $1 \neq n \in N$ and suppose $1 \neq x \in H \cap n H n^{- 1}$ . Then $y : = n^{- 1} x n$ satisfies $y \in H$ and $x = n y n^{- 1}$ , so $x y^{- 1} = x n^{- 1} x^{- 1} n$ . Here $x n^{- 1} x^{- 1} \in N$ by [F2] and $n \in N$ , so $x y^{- 1} \in N$ ; as also $x y^{- 1} \in H$ , the triviality of $N \cap H$ forces $x y^{- 1} = 1$ , that is $y = x$ . Thus $n^{- 1} x n = x$ , i.e. $x n = n x$ and $x \cdot n = n$ : the nonidentity element $x \in H$ fixes the nonidentity element $n \in N$ .

F2F5assume-hypalgebra

2.1

In the situation of step 1.1 the element $n$ satisfies $n \notin H$ : otherwise $n \in N \cap H = {1}$ , contrary to $n \neq 1$ ; hence also $n^{- 1} \notin H$ . The complement condition [F3] therefore gives $H \cap n^{- 1} H n = {1}$ , so step 1.1 forces $h = 1$ , contradicting $h \neq 1$ . Hence no nonidentity $n \in N$ is fixed by a nonidentity $h \in H$ , which is claim 1.

F3F4step 1.1contradiction

2.2

Step 1.3 contradicts freeness of the action on $N ∖ {1}$ ; therefore $H \cap n H n^{- 1} = {1}$ for every $1 \neq n \in N$ . By step 1.2 every $g \notin H$ has $H \cap g H g^{- 1} = H \cap n H n^{- 1}$ for some $n \neq 1$ in $N$ , so $H \cap g H g^{- 1} = {1}$ for every $g \notin H$ .

step 1.2step 1.3contradiction

3.1

Finally $1 < H$ holds by hypothesis and $H \neq G$ : if $H = G$ then $N = N \cap G = N \cap H = {1}$ , contradicting $1 < N$ . Hence ${1} < H < G$ and $H \cap g H g^{- 1} = {1}$ for all $g \notin H$ , so $H$ is a Frobenius complement of $G$ by [F3], which is claim 2. ∎

F1F3step 2.2given

Depends on

Used by

Affine linear Frobenius groups over finite fields Example
S₃ as a Frobenius group Example

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alex Bartel, Introduction to Representation Theory of Finite Groups, §6.1 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)